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stealth61 [152]
4 years ago
7

Two trains on parallel tracks are traveling in the same direction. One train starts 10 km behind the other. It overtakes the fir

st train in 2 hours . What is the relative speed of the second train with respect to the first train?
Physics
1 answer:
aleksklad [387]4 years ago
4 0

Answer:

v'=v_2-v_1=5\frac{km}{h}

Explanation:

The relative speed between the trains is given by:

v'=v_2-v_1

v1: speed of the first train

v2: speed of the second train

To find the relative velocity you write the equation of motion of both train by taking into account the information of the statement. The trains have constant velocity. Furthermore, the second train stars 10 km behind the other one. hence you have:

x=10+v_1t\\\\x=v_2t

for t = 2 hours, the positions of both trains are equal, thus you have:

10+v_1t=v_2t

you factor v2 - v1 from the last equation and replace t = 2h:

10=v_2t-v_1t\\\\10=(v_2-v_1)t\\\\v_2-v_1=\frac{10}{t}=\frac{10}{2}=5 \frac{km}{h}

Hence, you have that the relative velocity is:

v'=v_2-v_1=5\frac{km}{h}

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Read 2 more answers
Please help me
qaws [65]

-- We know that the y-component of acceleration is the derivative of the
y-component of velocity.

-- We know that the y-component of velocity is the derivative of the
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-- We're given the y-component of position as a function of time.

So, finding the velocity and acceleration is simply a matter of differentiating
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Now, the position function may look big and ugly in the picture.  But with the
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From the picture . . . . . y (t) = (1/2) (a₀ - g) t² - (a₀ / 30t₀⁴ ) t⁶

First derivative . . . y' (t) = (a₀ - g) t  -  6 (a₀ / 30t₀⁴ ) t⁵  =  (a₀ - g) t  -  (a₀ / 5t₀⁴ ) t⁵

There's your velocity . . . /\ .

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Pick, or look up, some reasonable figures for a₀ and t₀
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The big name in model rocketry is Estes.  Their website will give you
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